Question

A cold roll aluminium strip of 1400 mm width from 4 mm to 3.2 mm thickness on a 4 high mill. The diameter of the roll is 450 mm. Plane strain of aluminium at inlet as used is 0.5 and corresponding plane stress is $135N/m{m}^2$. Take the value of $\mu$ as 0.05 and strain at the end of rolling as 0.8 when plane stress is $148N/m{m}^2$. The expression of roll force is given by

$F=\omega \sqrt{R\Delta h}{\sigma }_m(1+\frac{\mu \sqrt{R\Delta h}}{h_b+h_f})N$

where symbols carry usual meaning.

The rolling load will be_____tonne.

• A. 296.168

Answer 1

The correct option is A 296.168
Given $\omega =1400mm,h_i=4mm,h_f=3.2mm$

${\sigma }_m=\frac{135+148}{2}=141.5N/m{m}^2$

Roll force $=1400\sqrt{\frac{450}{2}\left(0.8\right)}\times 141.5(1+\frac{0.05\sqrt{\frac{450}{2}\left(0.8\right)}}{4.0+3.2})$

$=2905415.398N$

$=296.168tonne$